Calculus Problem

Q:
A tank is leaking at a velocity of (6.5 - 0.16x^2) dm^3/h where x is time in hours from 12.00 when it was 20 dm^3 in the tank.
a) How much is left 15.00?
b) When is the tank empty?
A:
a. Integral(from 0 to 3)[6.5-0.16x^2] = F(3) - F(0) = 6.5 * 3 - 0.16 * 3^3 / 3 - 0 ~= 18
Answer: ~ 2 dm^3
b. Neeed haalp!

9.

>>1
What's 18 + 2, dude?
20. Coincidence?
That integral is how much liquid has leaked from the tank. So you subtract it from the initial amount to get how much is left.

For b, set it up similarly.
20 - ∫(6.5 - .16t^2)dt [from 0 to x] = 0 and solve for x. As in, you're saying, "20 minus the amount leaked from the beginning to x hours leaves nothing in the tank." You'll get a cubic with one negative (and, of course, irrelevant) solution and two positive solutions. Obviously, the correct one is the positive one closest to zero. [PROTIP: It's around three-and-a-half.]

:D

I don't want to solve cubics

I thought for a second it said Cactus problem

there is a formula for solving cubics...

>>4
Oh, for fuck's sake, if you're in calculus, surely you can do synthetic freakin' division.